Question

A car starting from rest and moving with uniform acceleration possesses average velocity of 5m/s, 10m/s and 25m/s in the first, second and third seconds. What is the total distance covered by the car in these three seconds.

Answer 1

Hint: As a very first step, one could read the question properly and hence note down the given values. So, we have three velocities and we could find the average of the three. Now, you could recall the definition for velocity and hence its expression. Substituting these you will get the answer.
Formula used:
Distance travelled,
$D=vt$

Complete answer:
In the question, we are given a car that starts from rest and moves with certain uniform acceleration. We are being said that it travels with 5m/s, 10m/s and 25m/s velocities in the first, second and third seconds. We are supposed to find the total distance that is covered by the car during those three seconds.
This is a very basic question taken from the topic of one dimensional motion. As a first step, you could find the average of the velocities given. That is, we will add all three and divide the sum by 3.
${{v}_{avg}}=\dfrac{5+10+25}{3}=\dfrac{40}{3}$
Now, let us recall the definition for velocity as the time rate of change of displacement. So,
$v=\dfrac{D}{t}$
The distance covered in the given three seconds would be,
$D=vt=\dfrac{40}{3}\times 3$
$\therefore D=40m$
Therefore, we found the total distance covered by the car in the given three seconds to be 40m.

Note:
One dimensional motion is a pretty much easy topic that comes under kinematics. So, all you have to do here is understand the small definitions that come under this topic and you will be able to solve these types of problems very easily.